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(D) The number of $\alpha$-particles emitted is given by $n_{\alpha} = \frac{A - A'}{4} = \frac{200 - 168}{4} = \frac{32}{4} = 8$.The change in atomic

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$8$ and $6$

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(D) The number of $\alpha$-particles emitted is given by $n_{\alpha} = \frac{A - A'}{4} = \frac{200 - 168}{4} = \frac{32}{4} = 8$.The change in atomic number due to $8$ $\alpha$-particles is $8 	imes 2 = 16$. The new atomic number would be $90 - 16 = 74$.Since the final atomic number is $80$,the number of $\beta$-particles emitted is $n_{\beta} = Z' - (Z - 2n_{\alpha}) = 80 - (90 - 16) = 80 - 74 = 6$.Thus,the number of $\alpha$ and $\beta$ particles are $8$ and $6$ respectively.

In the given radioactive decay,the number of $\alpha$ and $\beta$ particles emitted are respectively: $_{90}X^{200} \rightarrow _{80}Y^{168}$

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